Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
//Problem C

import java.util.*;
import java.io.*;

public class Main {

    public static boolean done(int[][] d) {
        for (int i = 0; i < d.length; i++) {
            for (int j = 0; j < d[0].length; j++) {
                if (d[i][j] != 0) {
                    return false;
                }
            }
        }
        return true;
    }

    public static void row(int c, int[][] d) {
        for (int i = 0; i < d[c].length; i++) {
            d[c][i]--;
        }
    }

    public static void col(int c, int[][] d) {
        for (int i = 0; i < d.length; i++) {
            d[i][c]--;
        }
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String s = br.readLine();
        String[] tokens = s.split(" ");

        int N = Integer.parseInt(tokens[0]);
        int K = Integer.parseInt(tokens[1]);

        int[][] d = new int[N][K];
        for (int i = 0; i < N; i++) {
            StringTokenizer st = new StringTokenizer(br.readLine());
            for (int j = 0; j < K; j++) {
                d[i][j] = Integer.parseInt(st.nextToken());
            }
        }

        int[] rows = new int[N];
        for (int i = 0; i < rows.length; i++) {
            rows[i] = Integer.MAX_VALUE;
        }
        int[] col = new int[K];
        for (int i = 0; i < col.length; i++) {
            col[i] = Integer.MAX_VALUE;
        }

        for (int i = 0; i < d.length; i++) {
            for (int j = 0; j < d[0].length; j++) {
                rows[i] = Math.min(rows[i], d[i][j]);
                col[j] = Math.min(col[j], d[i][j]);
            }
        }

        LinkedList<String> ans = new LinkedList<String>();

        if (K > N) {
            for (int i = 0; i < rows.length; i++) {
                while (rows[i] > 0) {
                    row(i, d);
                    rows[i]--;
                    ans.add("row " + (i + 1));
                }
            }

            for (int i = 0; i < d.length; i++) {
                for (int j = 0; j < d[0].length; j++) {
                    rows[i] = Math.min(rows[i], d[i][j]);
                    col[j] = Math.min(col[j], d[i][j]);
                }
            }

            for (int i = 0; i < col.length; i++) {
                while (col[i] > 0) {
                    col(i, d);
                    col[i]--;
                    ans.add("col " + (i + 1));
                }
            }
        } else {
            for (int i = 0; i < col.length; i++) {
                while (col[i] > 0) {
                    col(i, d);
                    col[i]--;
                    ans.add("col " + (i + 1));
                }
            }

            for (int i = 0; i < d.length; i++) {
                for (int j = 0; j < d[0].length; j++) {
                    rows[i] = Math.min(rows[i], d[i][j]);
                    col[j] = Math.min(col[j], d[i][j]);
                }
            }
            
            for (int i = 0; i < rows.length; i++) {
                while (rows[i] > 0) {
                    row(i, d);
                    rows[i]--;
                    ans.add("row " + (i + 1));
                }
            }

        }

        if (!done(d)) {
            System.out.println("-1");
        } else {
            StringBuilder sb = new StringBuilder();
            sb.append(ans.size()).append("\n");
            for (String e : ans) {
                sb.append(e);
                sb.append("\n");
            }
            System.out.println(sb.toString());
        }

//        for (int i = 0; i < d.length; i++) {
//            System.out.println(Arrays.toString(d[i]));
//        }
    }
}